Identifying poles of $f(z) =\frac{z^2+1}{z^2-1}$, z = 1

Question

$$f(z) =\frac{z^2+1}{z^2-1},\; z = 1$$ I'm trying to find the type of pole or just the poles of this function and my first attempt was to split this apart and try to find a laurent series so it would split into $\left((z^2+1)\left(\frac{1}{1-(-z)}\right)\left(\frac{-1}{1-z}\right)\right)$ and then I would try to find the the coeffcient from the series. But this seems to take too long as I would have to multiply two series together and then multiply again by $(z^2+1)$. Is there a better way to do this? I had this on a test and the test expected you to do things quickly.

Answer 1

$$f(z)=\frac{(z+i)(z-i)}{(z+1)(z-1)}$$

so there are simple poles at $z=1,-1$.

Answer 2

This is a simple pole as $\lim_{z\to1}f(z)(z-1)$ exists and is nonzero.